js排序的时间复杂度_javascript的array.indexOf的时间复杂度
是多少？
到匹配未排序数组中的值的第⼀个索引的最有效的⽅法是按顺序遍历列表，即O(n).
MDN还有⼀些提⽰：
Returns the first index at which a given element can be found in the array, or -1 if it is not present.
[…]
fromIndex
Default: 0 (Entire array is searched)
The index to start the search at. If the index is greater than or equal to the array’s length, -1 is returned, which means the array will not be searched. If the provided index value is a negative number, it is taken as the offset from the end of the array. Note: if the provided index is negative, the array is still searched from front to back. If the calculated index is less than 0, then the whole array will be searched.
EncodedJSValue JSC_HOST_CALL arrayProtoFuncIndexOf(ExecState* exec)
{
// 15.4.4.14
JSObject* thisObj = exec->hostThisValue().toThis(exec, StrictMode).toObject(exec);
unsigned length = thisObj->get(exec, exec->propertyNames().length).toUInt32(exec);
if (exec->hadException())
indexof的用法javascriptreturn JSValue::encode(jsUndefined());
unsigned index = argumentClampedIndexFromStartOrEnd(exec, 1, length);
JSValue searchElement = exec->argument(0);
for (; index < length; ++index) {
JSValue e = getProperty(exec, thisObj, index);
if (exec->hadException())
return JSValue::encode(jsUndefined());
if (!e)
continue;
if (JSValue::strictEqual(exec, searchElement, e))
return JSValue::encode(jsNumber(index));
}
return JSValue::encode(jsNumber(-1));
}

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